The Unapologetic Mathematician

Mathematics for the interested outsider

An Alternate Approach to Integration

We can wrap up this introduction to the Lebesgue integral by outlining the alternate approach that commenter Cristi was referring to. We’ll do this from the perspective of our current track, and it should be clear how the alternative definitions would lead us to the same place.

If f is a non-negative integrable function on a measure space (X,\mathcal{S},\mu). For every measurable set E\in\mathcal{S}, we define

\displaystyle a(E)=\inf\limits_{x\in E}f(x)

Also, for every finite, pairwise disjoint collection \mathcal{E}=\{E_1,\dots,E_n\}\subseteq\mathcal{S} of measurable sets we define

\displaystyle s(\mathcal{E})=\sum\limits_{i=1}^na(E_i)\mu(E_i)

We then assert that the supremum of all numbers s(\mathcal{E}) for all finite, pairwise disjoint collections \mathcal{E}\subseteq\mathcal{S} is equal to the integral of f:

\displaystyle\int f\,d\mu=\sup s(\mathcal{E})

If f is simple, this is obvious. Indeed, if \mathcal{E} is the collection of sets used to write f as a finite linear combination of characteristic functions, then s(\mathcal{E}) is exactly the integral of f by definition. Any set E that extends outside one of these sets will have a(E)=0, and so we can’t get any larger than the integral of f.

On the other hand, for a general integrable function f we consider a non-negative simple g with g\leq f, and we let \mathcal{E} be the sets used to express g as a finite linear combination of characteristic functions:

\displaystyle g=\sum\limits_{i=1}^n\alpha_i\chi_{E_i}

We see that

\displaystyle\int g\,d\mu=\sum\limits_{i=1}^n\alpha_i\mu(E_i)\leq\sum\limits_{i=1}^n a(E_i)\mu(E_i)=s(\mathcal{E})

If \{g_n\} is an increasing sequence of non-negative simple functions converging pointwise a.e. to f, then

\displaystyle\int f\,d\mu\lim\limits_{n\to\infty}\int g_n\,d\mu\leq\sup s(\mathcal{E})

where we use the definition of integrability, and we take the supremum over finite, pairwise disjoint collections \mathcal{E}. But it’s also clear that for every \mathcal{E} we have

\displaystyle s(\mathcal{E})=\int h\,d\mu\leq\int f\,d\mu

for some non-negative simple h\leq f.

So the alternate approach proceeds by defining the integral of a simple function as before, and defining general integrals of non-negative functions by the supremum above. General integrable functions overall are handled by using their positive and negative parts. Then you can prove the monotone convergence theorem, followed by Fatou’s lemma, and then the Fatou-Lebesgue theorem, which leads to dominated convergence theorem, and we’re pretty much back where we started.

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June 18, 2010 - Posted by | Analysis, Measure Theory

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